A 0.300-kg puck, initially at rest on a horizontal, frictionless surface, is str

Question

A 0.300-kg puck, initially at rest on a horizontal, frictionless surface, is struck by a 0.200-kg puck moving initially along the x axis with a speed of 2.00 m/s. After the collision, the 0.200-kg puck has a speed of 1.00 m/s at an angle or ?-50.0° to the positive x axis (see the figure below) Before the collision ni mi2 After the collision Ulf cos ? Jo cos ? hy sin (a) Determine the velocity of the 0.300-kg puck after the collision. m/s b) Find the fraction of kinetic energy transferred away or transformed to other forms of energy in the collision

Explanation / Answer

Solution:

pix = (0.20 kg)(2.0 m/s) = 0.40 kg m/s

piy = 0

After collision:

pfx= (0.20 kg)(1.0 m/s)(cos 50) + (0.30 kg)(vx)

= 0.129 + 0.30 vx

pfy= (0.20 kg)(1.0 m/s)(sin 50) + (0.30 kg)(vy)

= 0.153 + 0.30 vy

Conservation of momentum:

pix = pfx

0.40 kg m/s = 0.129 + 0.30 vx

vx= (0.40 - 0.129)/0.30 = 0.90 m/s

piy = pfy

0 = 0.153 + 0.30 vy

vy= -0.51 m/s

v = [(vx)2 + (vy)2]1/2 = 1.03 m/s

(b) Ki = (1/2)mv2 = (0.5)(0.20 kg)(2.0 m/s)2 = 0.40 J

Kf = (1/2)mv12 + (1/2)mv22 = 0.10 J + 0.16 J = 0.26 J

Ki - Kf = 0.14

Fraction = (0.14)/(0.40) = 0.35

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